Problem 19
Question
A car weighing \(1400 \mathrm{~kg}\) is moving at a speed of \(54 \mathrm{kmh}^{-1}\) up a hill when the motor stops. If it is just able to reach the destination which is at a height of \(10 \mathrm{~m}\) above the point, then the work done against friction (negative of the work done by the friction) is (Take \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(10 \mathrm{~kJ}\) (b) \(15 \mathrm{~kJ}\) (c) \(17.5 \mathrm{~kJ}\) (d) \(25 \mathrm{~kJ}\)
Step-by-Step Solution
Verified Answer
The work done against friction is 17.5 kJ.
1Step 1: Convert the Speed
Convert the speed of the car from km/h to m/s. The car's speed is given as 54 km/h.To convert km/h to m/s, use the formula: \[\text{Speed in } \mathrm{m/s} = \text{Speed in } \mathrm{km/h} \times \frac{5}{18}\]So, the speed is:\[54 \times \frac{5}{18} = 15 \mathrm{~m/s}\]
2Step 2: Calculate Initial Kinetic Energy
The kinetic energy (KE) of the car when it starts moving up the hill can be calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]Where:- \(m = 1400 \mathrm{~kg}\) is the mass of the car,- \(v = 15 \mathrm{~m/s}\) is the speed of the car.Substitute the values:\[ KE = \frac{1}{2} \times 1400 \times (15)^2 = 157500 \mathrm{~J} \]
3Step 3: Calculate Potential Energy at Height
Calculate the gravitational potential energy (PE) at the height of 10 meters using the formula:\[ PE = mgh \]Where:- \(m = 1400 \mathrm{~kg}\),- \(g = 10 \mathrm{~ms}^{-2}\),- \(h = 10 \mathrm{~m}\).Substitute the values:\[ PE = 1400 \times 10 \times 10 = 140000 \mathrm{~J} \]
4Step 4: Determine the Work Done Against Friction
The work done against friction is the difference between the initial kinetic energy and the gravitational potential energy at the height. It can be calculated as:\[ \text{Work Done (against friction)} = KE - PE \]Substitute the values from Step 2 and Step 3:\[ 157500 - 140000 = 17500 \mathrm{~J} \]Convert \(17500 \mathrm{~J}\) to \(17.5 \mathrm{~kJ}\).
Key Concepts
Kinetic EnergyGravitational Potential EnergyConversion of Units
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It depends on two main factors: the mass of the object (\( m \)) and the speed or velocity (\( v \)) at which the object is moving. The formula for calculating kinetic energy is \( KE = \frac{1}{2}mv^2 \).
Understanding kinetic energy helps us see how much energy is available for a moving object to perform work, such as moving against friction or overcoming gravity.
- Here, \( m \) represents the mass of the car, measured in kilograms (kg).
- \( v \) is the velocity of the car in meters per second (m/s).
Understanding kinetic energy helps us see how much energy is available for a moving object to perform work, such as moving against friction or overcoming gravity.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy that an object possesses due to its position in a gravitational field. This type of energy is associated with height, so when an object is lifted to a higher position, its potential energy increases. The formula used to calculate gravitational potential energy is \( PE = mgh \).
- \( m \) stands for the mass of the object, which is the car in this case, expressed in kilograms (kg).
- \( g \) is the acceleration due to gravity, typically \( 9.8 \text{ m/s}^2 \), but here approximated to \( 10 \text{ m/s}^2 \).
- \( h \) signifies the height the object has reached, measured in meters (m).
Conversion of Units
Conversion of units is an essential skill in physics, allowing us to express measurements in different scales for consistency in calculations. In this exercise, the car's speed needed to be converted from kilometers per hour (km/h) to meters per second (m/s), a common practice to align with standardized units for calculations involving kinetic energy.
The conversion is done using the factor \( \frac{5}{18} \). Simply, you multiply the speed in km/h by this factor to convert it to m/s. For example, 54 km/h \( \times \frac{5}{18} = 15 \text{ m/s} \).
The conversion is done using the factor \( \frac{5}{18} \). Simply, you multiply the speed in km/h by this factor to convert it to m/s. For example, 54 km/h \( \times \frac{5}{18} = 15 \text{ m/s} \).
- Why do we use \( \frac{5}{18} \)? Because the number 5/18 is derived from the conversion factors of 1000 meters in a kilometer, and 3600 seconds in an hour. This makes calculations consistent and accurate within the metric system.
Other exercises in this chapter
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